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3 years ago

Determine the molarity of 56.00 grams of aluminum nitrate in 450 milliliters of solution?

Chemistry

1 answer:

SVETLANKA909090 [29]3 years ago

3 0

Answer:

4.62 M

Explanation:

Molarity = moles/volumes (L), so you need to find the moles and the volumes in liters.

Finding the volume is easy because you just have to convert mL to L, so the volume is 0.45 L

Next, find the moles. You can do this by using the molar mass of aluminum to convert the grams to moles. The molar mass of aluminum is 26.98 g/mol.

56 g * (1 mol/26.98 g) = 2.08 mol

Now, divide the moles (2.08) by the volume (.45 L)

Molarity = 4.62 M

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Answer:

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3 years ago

A 32.5 g iron rod, initially at 22.4 ∘C, is submerged into an unknown mass of water at 63.0 ∘C, in an insulated container. The f

Allisa [31]

Answer:

The mass of water $m_{w}$ = 39.18 gm

Explanation:

Mass of iron $m_{iron}$ = 32.5 gm

Initial temperature of iron $T_{1}$ = 22.4°c = 295.4 K

Specific heat of iron $C_{iron}$ = 0.448 $\frac{KJ}{kg K}$

Mass of water = $m_{w}$

Specific heat of water $C_{w} = 4.2 \frac{KJ}{kg K}$

Initial temperature of water $T_{2}$ = 336 K

Final temperature after equilibrium $T_{f}$ = 59.7°c = 332.7 K

When iron rod is submerged into water then

Heat lost by water = Heat gain by iron rod

$m_{w}$ $C_{w}$ ( $T_{2}$ - $T_{f}$ ) = $m_{iron}$ $C_{iron}$ ( $T_{f}$ - $T_{1}$ )

Put all the values in above formula we get

$m_{w}$ × 4.2 × ( 336 - 332.7 ) = 32.5 × 0.448 × ( 332.7 - 295.4 )

$m_{w}$ = 39.18 gm

Therefore the mass of water $m_{w}$ = 39.18 gm

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Answer:

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